Solve (a+3)^2+6=(a^2+2)-5

$$( a + 3 ) ^ { 2 } + 6 = ( a ^ { 2 } + 2 ) - 5$$

Solve for a

$$a=-3$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+3\right)^{2}.

$$a^{2}+6a+9+6=a^{2}+2-5$$

Add 9 and 6 to get 15.

$$a^{2}+6a+15=a^{2}+2-5$$

Subtract 5 from 2 to get -3.

$$a^{2}+6a+15=a^{2}-3$$

Subtract a^{2} from both sides.

$$a^{2}+6a+15-a^{2}=-3$$

Combine a^{2} and -a^{2} to get 0.

$$6a+15=-3$$

Subtract 15 from both sides.

$$6a=-3-15$$

Subtract 15 from -3 to get -18.

$$6a=-18$$

Divide both sides by 6.

$$a=\frac{-18}{6}$$

Divide -18 by 6 to get -3.

$$a=-3$$